ISSN 1 746-7233, England, UK World Journal of Modelling and Simulation Vol. 7 (2011) No. 1, pp. 52-57 Fibonacci tan-sec method for construction solitary wave solution to differential-difference equations Xinghua Fan, Shouxiang Yang, Dan Zhao, Guoliang Cai Faculty of Science, Jiangsu University, Jiangsu 212013, P. R. China (Received September 30 2009, Accepted March 31 2010) Abstract. The Fibonacci secant function is constructed in this paper. Fibonacci tan-sec method is presented to search for the exact traveling wave solutions of nonlinear differential- difference equations (DDEs). This method is essentially equivalent to the classical hyperbolic method. The discrete mkdv lattice and some other lattice equations are chosen to illustrate the efficiency and effectiveness of the Fibonacci tan-sec method. Keywords: Fibonacci secant function, differential-difference equation, explicit solution, mkdv lattice 1 Introduction Differential-difference equations (DDEs) are one of the focuses of nonlinear study [10]. DDEs play an important role in the modeling of many physical phenomena as the particle vibration in lattice, the current in electrical network, the pulse in biological chain. DDEs are also of importance in numerical simulation of nonlinear partial differential equations (NPDEs), the queuing problem, discretization in solid state and quantum physics. Unlike nonlinear partial differential equations which are fully continuous, there are more difficulties in finding the exact solutions for DDEs because DDEs are semi-discretized with some (or all) of their spacial variables discretized while time is usually kept continuous. Little work is being done to symbolically compute exact solutions of DDEs although there has been considerable work done on finding exact solutions to PDEs due to the availability of computer symbolic like Maple or Mathematica. There are many direct methods to construct exact solutions of PDEs [8]. One of the most effective is the hyperbolic function expansion method [1]. Much extension of the continuous hyperbolic function expansion method has been done in the recent years [2, 6, 13, 21]. Baldwin et al. [1] presented an algorithm to find exact travelling wave solutions of DDEs in terms of tanh function and found kink-type solutions in several spatially discrete nonlinear models such as the celebrated Toda lattice and the AblowitzCLadik lattice. Since then, many modified method have appeared [3, 14 20]. A modified F-function method [7] and Sine-Gordon expansion method [5] to construct new exact solutions of the general discrete Toda and mkdv equation were given by Fan. Ma et al. [9] constructed a symmetrical Fibonacci tane according to the symmetrical Fibonacci sine and cosine and devised an algorithm to obtain exact traveling wave solutions to the (2 + 1)-dimensional Toda lattice, the discrete nonlinear Schrödinger equation and a generalized Toda lattice. Zhu utilized the Exp-function method to construct two families of new generalized soliton solutions for the discrete (2 + 1)-dimensional Toda lattice equation [22]. Those works allow one to directly derive many different types of traveling wave solutions of DDEs. Research was supported by the Post-doctoral foundation of China (No. 20080441071), the Post-doctoral foundation of Jiangsu Province(No.0802073c) and the High-level Talented Person Special Subsidizes of Jiangsu University (No. 08JDG013). Corresponding author. Tel.: +86-13775548931. E-mail address: fan131@ujs.edu.cn. Published by World Academic Press, World Academic Union
World Journal of Modelling and Simulation, Vol. 7 (2011) No. 1, pp. 52-57 53 We are interested in finding new forms of solutions of DDEs. In our present paper, Fibonacci secant function is constructed. Then the discrete Fibonacci tan-sec expansion method is introduced and is applied to the discrete mkdv and other lattice equations. 2 Fibonacci secant function There is deep connection between the Golden Section, Fibonacci numbers, and hyperbolic functions [11, 12]. Based on the Golden Section, the symmetrical Fibonacci sine (sfs) and the symmetrical Fibonacci cosine (cfs) functions were defined in [11]: sf s(x) = τ x τ x 5, cf s(x) = τ x + τ x 5, (1) where τ = 1+ 5 2. Those functions are introduced to consider so-called symmetrical representation of the hyperbolic Fibonacci functions and they may present a certain interest for modern theoretical physics taking into consideration a great role played by the Golden Section in modern physical researches. Ma et al. defined the Fibonacci tane function [9] tan F s(x) = τ x τ x τ x, (2) + τ x and used it to the (2 + 1)-dimensional Toda lattice, the discrete nonlinear Schrödinger equation and a generalized Toda lattice, and successfully constructed some explicit and exact traveling wave solutions. We define the Fibonacci secant function With Eqs. (1) (3) we have sec F s(x) = [cf s(x)] 2 [sf s(x)] 2 = 4 5, [sec F s(x)]2 = 4 5 (1 [tan F s(x)]2 ), d tan F s(x) dx tan F s(x ± y) = 1 cf s(x). (3) = ln(τ) (1 [tan F s(x)] 2 d sec F s(x) ), = ln(τ) sec F s(x) tan F s(x), dx tan F s(x) ± tan F s(y) 1 ± tan F s(x) tan F s(y), sec F s(x ± y) = 2 sec F s(x) sec F s(y) 5 1 ± tan F s(x) tan F s(y). (4) These functions are too similar to the classical hyperbolic functions but differ from them by one peculiarity. In contrast to the classical hyperbolic functions the Fibonacci hyperbolic functions have a numerical analogy, Fibonacci numbers [11]. Those properties in Eq. (4) play a crucial role in our Fibonacci tan-sec method. 3 Discrete Fibonacci tan-sec method In this section, the discrete Fibonacci tan-sec method for construction explicit solitary wave solutions to DDEs is demonstrated. The main point of discrete Fibonacci tan-sec method is to suppose the solution of DDEs be a combination of the Fibonacci tane and Fibonacci secant functions. There are mainly five steps. For a given system of M polynomial DDEs (u n+p1 (X),u n+p2 (X),, u n+pk (X), u n+p 1 (X), u n+p 2 (X),..., u n+p k (X),, u (r) n+p 1 (X), u (r) n+p 2 (X),, u (r) n+p k (X)) = 0, (5) WJMS email for subscription: info@wjms.org.uk
54 X. Fan & S. Yang & et al.: Fibonacci tan-sec method for construction solitary wave solution where the dependent variable u n and its shift have M components u i,n, the continuous variable X has N components x i, the discrete variable n has Q components, the k shift vectors p i Z Q, and U (r) (X) denotes the collection of mixed derivative terms of order r. The Greek letters represent positive parameters of the system. Step 1. Travelling wave transformation. We seek for the following formal travelling wave solutions which are of important physical significance. Let ξ n = Q N d i n i + c j x i + δ = d n + c x + δ, (6) i=1 j=1 where coefficients c 1, c 2,, c N, d 1, d 2,, d Q and the initial phase δ are constants. The dot presents the Euclidean inner product. Then Eq. (5) is turned to (u n+p1 (ξ n ),u n+p2 (ξ n ),, u n+pk (ξ n ), u n+p 1 (ξ n ), u n+p 2 (ξ n ),, u n+p k (ξ n ),, Step 2. Solution assumption. We assume solution to Eq. (7) be where u (r) n+p 1 (ξ n ), u (r) n+p 2 (ξ n ),, u (r) n+p k (ξ n ) = 0. (7) M i U i,n (ξ n ) = (a ij Tn j + B ij Tn j 1 S n ), (8) j=0 T n = tan F s(ξ n ); S n = sec F s(ξ n ), (9) and a ij, b ij are constants to be determined later. The constant M i can be decided by balancing the highest the highest-order derivative term with nonlinear terms [4]. Step 3. Derive the algebraic system. Substituting Eq. (8) into Eq. (5), collecting the coefficients of T i ns j n(i = 0, 1,, j = 0, 1) and setting them to zero, we can obtain an nonlinear algebra system with respect to a ij, b ij. Step 4. Solve the algebraic system. Solving the algebraic system, we may find the values of unknowns. Step 5. Build and test the exact wave solutions. Substituting the algebraic solutions gotten in Step 3, Eq. (6) and Eq. (9) into Eq. (8), we can obtain the traveling wave solutions of nonlinear DDE (Eq. (5)). 4 Applications of the discrete Fibonacci tan-sec method In this section, we apply the discrete Fibonacci tan-sec method to some discrete lattice equations. 4.1 Discrete mkdv lattice Consider the following discrete mkdv lattice equation du n dt = (α u 2 n)(u n+1 u n 1 ), (10) where α is a constant. A kink solution was given by Baldwin et al. [1] and double period solutions by Dai [3]. First we take the wave transformation ξ = dn + ct + ξ 0, (11) WJMS email for contribution: submit@wjms.org.uk
World Journal of Modelling and Simulation, Vol. 7 (2011) No. 1, pp. 52-57 55 where d, c, ξ 0 are constants. Then Eq. (10) turns to c du n(ξ) dξ = (α u 2 n(ξ))(u n+1 (ξ) u n 1 (ξ)). (12) Analyzing the leading term and balancing du n (ξ)/dξ and u 2 n(ξ), we get M = 1. So we can assume that Eq. (12) has solutions as where A 0, A 1, B 1 are constants. Then u n (ξ) = A 0 + A 1 tan F s(ξ) + B 1 sec F s(ξ) A 0 + A 1 T F + B 1 S F, (13) du n (ξ) dξ = A 1 ln τ(1 T 2 F ) + B 1 ln τt F S F, (14) u n 1 (ξ) = A 0 + A 1 T F tan F s(d) 1 T F tan F s(d) + B 1 u n+1 (ξ) = A 0 + A 1 T F + tan F s(d) 1 + T F tan F s(d) + B 1 2 S F sec F s(d) 5 1 T F tan F s(d), (15) 2 S F sec F s(d) 5 1 + T F tan F s(d). (16) Substituting Eqs. (13) (16) into Eq. (12), we get an equation with respect to TF i Sj F (i = 0, 1,, j = 0, 1). Setting the coefficients be zero, we get a nonlinear algebraic system with respect to A 0, A 1, B 1, d, c, 10 tan F s(d)αa 1 10 tan F s(d)a 2 0A 1 25 2 tan F s(d)b2 1A 1 5c ln(τ)a 1 = 0, 20tan F s (d) A 0 B 1 A 1 = 0, 20tan F s (d) A 0 A 1 2 10tan F s (d) A 0 B 1 2 5sec F s (d) = 0, 4tan F s (d) A 1 2 B 1 5sec F s (d) + 20tan F s (d) A1 2 B 1 5tan F s (d) B 1 3 5sec F s (d) + 5c ln (τ) B 1 (tan F s (d)) 2 = 0, 25 2 tan F s (d) B 1 2 A 1 5c ln (τ) A 1 (tan F s (d)) 2 + 10tan F s (d) A 1 3 10tan F s (d) A 1 B 1 2 5sec F s (d) = 0, 10tan F s (d) A 0 B 1 2 5sec F s (d) 20tan F s (d) A 0 A 1 2 = 0, 5c ln (τ) B 1 + 5tan F s (d) B 1 3 5sec F s (d) + 4tan F s (d) A 0 2 B 1 5sec F s (d) 20tan F s (d) A 1 2 B 1 4tan F s (d) αb 1 5sec F s (d) = 0, 10tan F s (d) A 0 2 A 1 + 5c ln (τ) A 1 + 5c ln (τ) A 1 (tan F s (d)) 2 10tan F s (d) αa 1 10tan F s (d) A 1 3 + 10tan F s (d) A 1 B 1 2 5sec F s (d) + 25tan F s (d) B 1 2 A 1 = 0, 8tan F s (d) A 0 A 1 B 1 5sec F s (d) + 20tan F s (d) A0 B 1 A 1 = 0, With the help of Maple and Wu method we get two solutions of the algebraic system. Case 1. Case 2. where d is an arbitrary constant. A 0 = 0, A 1 = ± α tan F s(d), B 1 = 0, c = 2 α tan F s (d). (17) ln (τ) A 0 = 0, A 1 = 0, B 1 = ± 5α αsf s(d)i, c = tan F s(d), (18) ln(τ) WJMS email for subscription: info@wjms.org.uk
56 X. Fan & S. Yang & et al.: Fibonacci tan-sec method for construction solitary wave solution Inserting Eq. (17) and Eq. (18) into Eq. (13), we get the following solitary solutions to Eq. (10): u n (t) = ± α tan F s (d) α tan F s(d) tan F s(dn + 2 t + ξ 0 ), ln (τ) u n (t) = ± 5α αsf s(d)i sec F s(dn ln(τ) tan F s(d) + ξ 0). (19) We can see that solution (19) is a new solution to our knowledge. Now we apply the Fibonaccitan-sec method to some discrete nonlinear lattice equations. For simplification we only give the lattice equations and the solutions. 4.2 Hybrid lattice Hybrid lattice [1] has solutions u n (t) = 1 2 dun(t) dt α β + = (1 + αu n (t) + βu 2 n(t))(u n+1 (t) u n 1 (t)) ( 4β α 2 5 sf s(d) sec F s dn + 2β 4 ) α 2 4β β ln(τ) tan F s(d)t + ξ 0 and u n (t) = 1 2 α β + ( α 2 4β tan F s(d) tan F s dn + 1 2β 2 tan F s (d) ( 4 β + α 2) ) t + ξ 0, β ln (τ) where d, ξ 0 are arbitrary constants. 4.3 (2+1) dimensional Toda lattice (2+1) dimensional Toda lattice [1] 2 u n (x, t) x t ( un (x, t) = t ) + 1 (u n 1 (x, t) 2u n (x, t) + u n+1 (x, t)) has solution as u n = a 0 + 5sF s2 (d) 4 ln(τ) ( 1 tan F s c 2 dn + 5sF s2 (d) 4 ln 2 (τ) 1 x + c 2 t + ξ 0 c 2 ), where A 0, d, c 2, ξ 0 are arbitrary constants. 4.4 General Toda lattice General Toda lattice [1] has solution u n = du n dt = v n (u n+1 v n ) v n 1 (u n 1 v n 1 ), dv n dt = v n(u n+1 u n ) ln(τ) tan F s(d) c + ln(τ)c tan F s(dn + ct + ξ 0), v n = ln(τ) tan F s(d) c + ln(τ)c tan F s(dn + ct + ξ 0), where d, c, ξ 0 are arbitrary constants. We state that all those solutions are first given in the Fibonacci function forms. WJMS email for contribution: submit@wjms.org.uk
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